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I want to show that for a fixed $q$ the Beta function $B(p,q)$ is log convex. I tried to show that the second derivative of the Beta function with respect to $p$ is larger than zero, i.e. that:

\begin{equation} \partial_p^2 \log(\int_0^1t^{p-1}(1-t)^{q-1} dt) = \partial_p(\frac{\int_0^1t^{p-1}\log(t)(1-t)^{q-1} dt}{\int_0^1t^{p-1}(1-t)^{q-1} dt}) = \\ \frac{(\int_0^1t^{p-1}\log^2(t)(1-t)^{q-1} dt)(\int_0^1t^{p-1}(1-t)^{q-1} dt) - (\int_0^1t^{p-1}\log(t)(1-t)^{q-1} dt)^2}{(\int_0^1t^{p-1}(1-t)^{q-1} dt)^2} \geq 0 \end{equation} where I used the Leibniz integral rule twice.

So I need to show that \begin{equation} (\int_0^1t^{p-1}\log^2(t)(1-t)^{q-1} dt)(\int_0^1t^{p-1}(1-t)^{q-1} dt) - (\int_0^1t^{p-1}\log(t)(1-t)^{q-1} dt)^2 \geq 0 \end{equation}

But I am unsure on how to proceed - I don't know how to get rid of the integrals in this expression. Could you please help me?

Abby Jean
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  • Try Cauchy-Schwarz – Deane Aug 15 '21 at 20:38
  • For any measure space $(A,\mathscr{F},\mu)$ and measurable function $h\geq0$, the set $E={p\geq-:\int_X h^p,d\mu<\infty}$ is convex. If $E$ has interior then the map $p\mapsto \log\Big(\int_Xh^p,d\mu\Big)$ is convex. See. I your case $X=[0,1]$, $\mathscr{F}$ is the Borel $\sigma$-algebra, and $\mu(dx)=x(1-x)^{1-p},dx$. – Mittens Aug 15 '21 at 21:07

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